How many bacteria were present when t=2?

A bacteria culture starts with 500 bacteria and doubles in size every half hour.1

(a) How many bacteria are there after 3 hours?

We are told “. . . doubles in size every half hour.” Let’s make a table of the time and population:

t 0 0.5 1.0 1.5 2.0 2.5 3.0

bacteria 500 1000 2000 4000 8000 16000 32000

Thus, after three hours, the population of bacteria is 32,000.

(b) How many bacteria are there after t hours?

In t hours, there are 2t doubling periods. (For example, after 4 hours, the population has doubled 8

times.) The initial value is 500, so the population P at time t is given by

P(t) = 500 · 2

2t

This is an acceptable response, but in calculus and all advanced mathematics and science, we will

almost always want to use the natural exponential base, e. Let’s redo the problem using the natural

exponential growth function

P(t) = P0e

rt

We are given that the initial population is 500 bacteria. So P0 = 500 and we have

P(t) = 500ert

We know that after 1 hour, there are 2000 bacteria. (We could’ve used any other pair from our table

that we wish.) Substituting into our function, we get

P(t) = 500ert

2000 = 500er·1

2000 = 500er

1

500

· 2000 =

1

500

· 500er

4 = er

and now we use the natural logarithm to solve for r

ln (4) = ln (er

)

ln (4) = r · ln (e)

ln (4) = r

1.3863 ≈ r

Thus the function P(t) = 500e1.3863t gives the number of bacteria after t hours.

1Stewart, Calculus, Early Transcendentals, p. 54, #30.

Step-by-step explanation:

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

ljn

step-by-step explanation: